To find the relative molecular mass of gas R using Graham's law of diffusion, we can use the following relationship:
\[ \frac{t_1}{t_2} = \sqrt{\frac{M_2}{M_1}} \]
where:
- \( t_1 \) and \( t_2 \) are the times taken for the two gases to diffuse,
- \( M_1 \) and \( M_2 \) are the molar masses of the gases (in this case, oxygen and gas R, respectively).
Let:
- \( t_1 = 20 \) seconds (for oxygen),
- \( t_2 = 30 \) seconds (for gas R),
- \( M_1 = 32 \) g/mol (molar mass of oxygen, \( O_2 \)),
We need to find \( M_2 \) (the molar mass of gas R).
Plugging the values into the equation gives us:
\[ \frac{20}{30} = \sqrt{\frac{M_2}{32}} \]
Simplifying the fraction on the left side:
\[ \frac{2}{3} = \sqrt{\frac{M_2}{32}} \]
Now, squaring both sides to eliminate the square root gives:
\[ \left(\frac{2}{3}\right)^2 = \frac{M_2}{32} \]
Calculating the left side:
\[ \frac{4}{9} = \frac{M_2}{32} \]
Now, to find \( M_2 \), we cross-multiply:
\[ 4 \cdot 32 = 9 \cdot M_2 \]
\[ 128 = 9 M_2 \]
Now, divide both sides by 9:
\[ M_2 = \frac{128}{9} \approx 14.22 \text{ g/mol} \]
Thus, the relative molecular mass of gas R is approximately 14.22 g/mol.
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